Problem: Is ${143304}$ divisible by $3$ ?
Explanation: A number is divisible by $3$ if the sum of its digits is divisible by $3$ . [ Why? First, we can break the number up by place value: $ \begin{eqnarray} {143304}= &&{1}\cdot100000+ \\&&{4}\cdot10000+ \\&&{3}\cdot1000+ \\&&{3}\cdot100+ \\&&{0}\cdot10+ \\&&{4}\cdot1 \end{eqnarray} $ Next, we can rewrite each of the place values as $1$ plus a bunch of $9$ s: $ \begin{eqnarray} {143304}= &&{1}(99999+1)+ \\&&{4}(9999+1)+ \\&&{3}(999+1)+ \\&&{3}(99+1)+ \\&&{0}(9+1)+ \\&&{4} \end{eqnarray} $ Now if we distribute and rearrange, we get this: $ \begin{eqnarray} {143304}= &&\gray{1\cdot99999}+ \\&&\gray{4\cdot9999}+ \\&&\gray{3\cdot999}+ \\&&\gray{3\cdot99}+ \\&&\gray{0\cdot9}+ \\&& {1}+{4}+{3}+{3}+{0}+{4} \end{eqnarray} $ Any number consisting only of $9$ s is a multiple of $3$ , so the first five terms must all be multiples of $3$ That means that to figure out whether the original number is divisible by $3 $ , all we need to do is add up the digits and see if the sum is divisible by $3$ . In other words, ${143304}$ is divisible by $3$ if ${ 1}+{4}+{3}+{3}+{0}+{4}$ is divisible by $3$ Add the digits of ${143304}$ $ {1}+{4}+{3}+{3}+{0}+{4} = {15} $ If ${15}$ is divisible by $3$ , then ${143304}$ must also be divisible by $3$ ${15}$ is divisible by $3$, therefore ${143304}$ must also be divisible by $3$.